\newproblem{lay:2_4_19}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.4.19}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into
	the following system of linear equations:
	\begin{center}
		$\begin{pmatrix}A-sI_n & B \\ C & I_m\end{pmatrix}\begin{pmatrix}\mathbf{x} \\ \mathbf{u}\end{pmatrix}=\begin{pmatrix}\mathbf{0} \\ \mathbf{y}\end{pmatrix}$
	\end{center}
	where $A$ is $n\times n$, $B$ is $n\times m$, $C$ is $m\times n$ and $s$ is a variable. The vector $\mathbf{u}\in\mathbb{R}^m$ is the ``input'' to the system,
	$\mathbf{y}\in \mathbb{R}^m$ is the ``output'' of the system, and $\mathbf{x}\in\mathbb{R}^n$ is the ``state'' vector. Actually, the vectors $\mathbf{u}$,
	$\mathbf{x}$ and $\mathbf{y}$ are functions of $s$, but this does not affect the algebraic calculations of this exercise.
	
	Assume $A-sI_n$ is invertible and view the previous equation as a system of two matrix equations. Solve the top equation for $\mathbf{x}$ and substitute in the
	bottom equation. The result is an equation of the form $W(s)\mathbf{u}=\mathbf{y}$, where $W(s)$ is a matrix that depends on $s$. $W(s)$ is called the
	\textit{transfer function} of the system because it transforms the input $\mathbf{u}$ into the output $\mathbf{y}$. Find $W(s)$ and describe how it is related
	to the partitioned system matrix of the equation above.
}{
  % Solution
	The first equation gives us
	\begin{center}
		$(A-sI_n)\mathbf{x}+B\mathbf{u}=\mathbf{0} \Rightarrow \mathbf{x}=-(A-sI_n)^{-1}B\mathbf{u}$
	\end{center}
	Now we go with the second equation and substitute this value into it
	\begin{center}
		$C\mathbf{x}+\mathbf{u}=\mathbf{y}$\\
		$C(-(A-sI_n)^{-1}B\mathbf{u})+\mathbf{u}=\mathbf{y}$ \\
		$(-C(A-sI_n)^{-1}B+I_m)\mathbf{u}=\mathbf{y}$ \\
		$(I_m-C(A-sI_n)^{-1}B)\mathbf{u}=\mathbf{y}$ \\
	\end{center}
	So, the transfer function is given by the matrix $W(s)=I_m-C(A-sI_n)^{-1}B$. 
}
\useproblem{lay:2_4_19}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
